Abstract
In this article, we discuss the existence and multiplicity of positive solutions for the sixthorder boundary value problem with three variable parameters as follows:
where A(t), B(t), C(t) ∈ C[0,1], f(t, u) : [0,1] × [0, ∞) → [0. ∞) is continuous. The proof of our main result is based upon spectral theory of operators and fixed point theorem in cone.
Keywords:
sixthorder differential equation; positive solution; fixed point theorem; spectral theory of operators1 Introduction
In this article, we study the existence and multiplicity of positive solution for the following nonlinear sixthorder boundary value problem (BVP for short) with three variable parameters
where A(t), B(t), C(t) ∈ C[0,1], f(t, u) : [0,1] × [0, ∞) → [0. ∞) is continuous.
In recent years, BVPs for sixthorder ordinary differential equations have been studied extensively, see [17] and the references therein. For example, Tersian and Chaparova [1] have studied the existence of positive solutions for the following systems (1.2):
where A, B, and C are some given real constants and f(x, u) is a continuous function on R^{2}, is motivated by the study for stationary solutions of the sixthorder parabolic differential equations
This equation arose in the formation of the spatial periodic patterns in bistable systems and is also a model for describing the behaviour of phase fronts in materials that are undergoing a transition between the liquid and solid state. When f(x, u) = u  u^{3}, it was studied by Gardner and Jones [2] as well as by Caginalp and Fife [3]. In [1], existence of nontrivial solutions for (1.2) is proved using a minimization theorem and a multiplicity result using Clarks theorem when C = 1 and f(x, u) = u^{3}. The authors have studied also the homoclinic solutions for (1.2) when C = 1 and f(x, u) = a(x)uu^{σ}, where a(x) is a positive periodic function and σ is a positive constant by the mountainpass theorem of BrezisNirenberg and concentrationcompactness arguments. In [4], by variational tools, including two BrezisNirenbergs linking theorems, Gyulov et al. have studied the existence and multiplicity of nontrivial solutions of BVP (1.2).
Recently, in [5], the existence and multiplicity of positive solutions of sixthorder BVP with three parameters
has been studied under the hypothesis of
(A_{1}) f : [0,1] × [0, ∞) → [0. ∞) is continuous.
(A_{2}) α, β, γ ∈ R and under the condition of satisfying
the existence and multiplicity for positive solution of BVP (1.3) are established by using fixed point index theory. In this article, we consider more general BVP (1.1), based upon spectral theory of operators and fixed point theorem in cone, we will establish the existence and multiplicity positive solution of BVP (1.1) and extend the result of [5] under appropriate conditions. Our ideas mainly come from [5,810].
We list the following conditions for convenience:
(H_{1}) f : [0,1] × [0, +∞) → [0. +∞) is continuous.
(H_{2}) A(t), B(t), C(t) ∈ C[0,1], α = min_{0≤t≤1 }A(t), β = min_{0≤t≤1 }B(t), γ = min_{0≤t≤1 }C(t), and satisfies
Let Y = C[0,1], Y_{+ }= {u ∈ Y : u(t) ≥ 0, t ∈ [0,1]}. It is well known that Y is a Banach space equipped with the norm u_{0 }= sup_{0≤t≤1 }u(t), u ∈ Y. Set X = { u ∈ C^{4}[0,1] : u(0) = u(1) = u''(0) = u''(1) = 0}, then X also is a Banach space equipped with the norm u_{X }= max {u(t)_{0}, u"(t)_{0}, u^{(4)}(t)_{0}}. If u ∈ C^{4}[0,1] ∩ C^{6}(0,1) fulfils BVP (1.1), then we call u is a solution of BVP (1.1). If u is a solution of BVP (1.1), and u(t) > 0, t ∈ (0, 1), then we say u is a positive solution of BVP (1.1).
2 Preliminaries
In this section, we will make some preliminaries which are needed to show our main results.
Lemma 2.1. Let u ∈ X, then u_{0 }≤ u"_{0 }≤ u^{(4)}_{0 }≤ u_{X}.
Proof. The proof is similar to the Lemma 1 in [8], so we omit it. □
Lemma 2.2. [5] Let λ_{1}, λ_{2}, and λ_{3 }be the roots of the polynomial P (λ) = λ^{3 }+ γλ^{2 } βλ + α. Suppose that condition (H_{2}) holds, then λ_{1}, λ_{2}, and λ_{3 }are real and greater than π^{2}.
Note : Based on Lemma 2.2, it is easy to learn that when the three parameters satisfy the condition of (H_{2}), they satisfy the condition of nonresonance.
Let G_{i}(t, s)(i = 1, 2, 3) be the Green's function of the linear BVP
u"(t) + λ_{i}u(t) = 0, u(0) = u(1) = 0,
Lemma 2.3. [10]G_{i}(t, s)(i = 1, 2, 3) has the following properties
(c1) G_{i}(t, s) > 0, ∀t, s ∈ (0, 1).
(c2) G_{i}(t, s) <C_{i}G_{i}(s, s), ∀t, s ∈ [0,1], in which C_{i }> 0 is constant.
(c3) G_{i}(t, s) ≥ δ_{i}G_{i}(t, t)G_{i}(s, s), ∀t, s ∈ [0,1], in which δ_{i }> 0 is constant.
We set
then starting from Lemma 2.3 we know M_{i}, m_{i}, C_{ij }> 0.
For any h ∈ Y, take into consideration of linear BVP:
where α, β, γ satisfy assumption (H_{2}). Since
then for any h ∈ Y, the LBVP(2.4) has a unique solution u, which we denoted by Ah = u. The operator A can be expressed by
Lemma 2.4. The linear operator A : Y → X is completely continuous and A ≤ ϖ, where ϖ = λ_{2}+λ_{3}(C_{1}C_{2}C_{3}M_{1}M_{2}M_{3}λ_{3}+C_{1}C_{2}M_{1}M_{2})+ λ_{2}λ_{3}(C_{1}C_{2}C_{3}M_{1}M_{2}M_{3}+C_{1}M_{1}).
Proof. It is easy to show that the operator A : Y → X is linear operator. ∀h ∈ Y, u = Ah ∈ X, u(0) = u(1) = u"(0) = u"(1) = u^{(4)}(0) = u^{(4)}(1) = 0. Let
by (2.5) and (2.7), we have
and
By (2.6), for any t ∈ [0,1], we have
Again, let ω = u" + λ_{3}u, then ω(0) = ω(1) = ω"(0) = ω"(1) = 0, by (2,5), we have
Then
So
Based on (2.8), (2.9), and (2.12), we have
where
So, u^{(4)}(t) ≤ ϖh_{0}, by Lemma2.1, u_{X }≤ ϖh_{0}, then
so A is continuous, and A ≤ ϖ.
Next, we will show that A is compact with respect to the norm ·_{X }on X.
Suppose {h_{n}}(n = 1, 2, . . .) an arbitrary bounded sequence in Y, then there exists K_{0 }>0 such that h_{n}_{0 }≤ K_{0}, n = 1, 2, . . . . Let u_{n }= Ah_{n}, 1, 2, ...By (2.8), ∀t_{1}, t_{2 }∈ [0, 1], t_{1 }< t_{2}, we have
Because G_{i}(t, s)(i = 1, 2, 3) is uniform continuity on [0,1] × [0,1], based on the above demonstration,
it is easy to proof that
The main tools of this article are the following wellknown fixed point index theorems.
Let E be a Banach Space and K ⊂ E be a closed convex cone in E. Assume that Ω is a bounded open subset of E with boundary ∂Ω, and K ∩ Ω ≠ ∅. Let
Let K_{r }= {u ∈ K u <r} and ∂K_{r }= {u ∈ K u <r} for every r >0.
Lemma 2.5. [12] Let A : K → K be a completely continuous mapping. If μAu ≠u for every u ∈ ∂K_{r }and 0 < μ ≤ 1, then i(A, K_{r}, K) = 1.
Lemma 2.6. [12] Let A : K → K be a completely continuous mapping. Suppose that the following two conditions are satisfied:
(i)
(ii) μAu ≠u for every u ∈ ∂K_{r }and μ ≥ 1,
then i(A, K_{r}, K) = 0.
Lemma 2.7. [12] Let X be a Banach space, and let K ⊆ X be a cone in X. For p >0, define
(i) If u ≤ Au, for every u ∈ ∂K_{p}, then i(A, K_{p}, K) = 0.
(ii) If u ≥ Au, for every u ∈ ∂K_{p}, then i(A, K_{p}, K) = 1.
3 Main results
We bring in following notations in this section:
Suppose that:
(H_{3}) L = ϖK <1, where ϖ is defined as in (2.14).
Theorem 3.1. Assume that (H_{1})(H_{3}) hold, and
(i)
Proof. ∀h ∈ Y, consider the LBVP
It is easy to prove (3.1) is equivalent to the following BVP
where Gv := (C(t)  γ)v^{(4) } (B(t) β)v" + (A(t)  α)v, ∀v ∈ X. Obviously, the operator G : X → Y is linear, and ∀v ∈ X, t ∈ [0,1], we have Gv(t) ≤ K v_{X}. Hence Gv_{0 }≤ K v_{X}, and so G ≤ K. On the other hand, u ∈ C^{4}[0,1]⋂C^{6}(0,1), t ∈ [0,1] is a solution of (3.2) iff u ∈ X satisfies u = A(Gu + h), i.e.,
Owing to G : X → Y and A : Y → X, the operator I  AG maps X into Y. From A ≤ ϖ (by Lemma 2.4) together with G ≤ K and condition (H_{3}), applying operator spectral theorem, we have that the operator (I  AG)^{1 }exists and is bounded. Let H = (I  AG)^{1}A, then (3.3) is equivalent to u = Hh. By the Neumann expansion formula, H can be expressed by
The complete continuity of A with the continuity of (I  AG)^{1 }yields that the operator H : Y → X is completely continuous. If we restrict H : Y_{+ }→ Y, ∀h ∈ Y_{+ }and mark u = Ah, then u ∈ X ∩ Y_{+}. Based on equation (2.8), (2.11) and Lemma 2.4, we have
by b(t) ≥ (λ_{2 }+ λ_{3})c(t) and
Hence
and so (AG)(Ah)(t) = A(GAh)(t) ≥ 0, ∀t ∈ [0,1]. Suppose that ∀h ∈ Y_{+}, (AG)^{k }(Ah)(t) ≥ 0, ∀t ∈ [0,1]. For any h ∈ Y_{+}, let h_{1 }= GAh, by (3.5) we have h_{1 }∈ Y_{+}, and so
Thus by induction it follows that ∀n ≥ 1, ∀h ∈ Y_{+}, (AG)^{n }(Ah)(t) ≥ 0, ∀t ∈ [0,1]. By (3.4), we have
So H : Y_{+ }→ Y_{+ }∩ X.
On the other hand, we have
So the following inequalities hold
For any u ∈ Y_{+}, define Fu = f(t, u). Based on condition (H_{1}), it is easy to show F : Y_{+ }→ Y_{+ }is continuous. By (3.1)(3.3), It is easy to see that u ∈ C^{4}[0,1] ∩ C^{6}(0, 1) is a positive solution of BVP (1.1) iff u ∈ Y_{+ }is a nonzero solution of an operator equation as follows
Let Q = HF. Obviously, Q : Y_{+ }→ Y_{+ }is completely continuous. We next show that the operator Q has at least one nonzero fixed point in Y_{+}.
Let
In which
Here M_{1 }and M_{2 }can be defined as that in (2.1), C_{12 }and C_{23 }can be defined as that in (2.2), C_{i},δ_{i}(i = 1, 2, 3) can be defined as that in Lemma 2.3. It is easy to prove that P is a cone in Y. We will prove QP ⊂ P next.
For any u ∈ P, let h = Fu, then h ∈ Y_{+}. By (3.6) and Lemma 2.3, we have
By Lemma 2.3, for all u ∈ P, we have
And accordingly we have
By using (c3) in Lemma 2.3, (3.8) and (3.11), we have
So
Let
in which m_{1 }can be defined as that in (2.1). It's easy to prove
Case (i), since
From (3.6) and (3.14), we have
Therefore,
After multiplying the two sides of the first equation in (3.15) by sin πt and integrating on [0,1], we have
then
Since
On the other hand, since
We choose
and so
Thus, by (3.19), we have
Then by Lemma 2.5 we know
Now, by the additivity of fixed point index, combine (3.17) and (3.20) to conclude that
Therefore Q has a fixed point in
Case (ii), since
Let r ∈ (0, ω), we now prove that μQu ≠ u for every u ∈ ∂Ω_{r}, and 0 < μ ≤ 1. In fact, suppose the contrary, then there exist u_{0 }∈ ∂Ω_{r}, and 0 < μ_{0 }≤ 1 such that μ_{0}Qu_{0 }= u_{0}. Let ω_{2 }= AFu_{0}, by (3.6), we have
Since
On the other hand, because
Let C = max_{0≤t≤1,0≤x≤H}f(t,x)  (Γ + ε)x + 1, then it is clear that
Choose R > R_{0 }= max {H/ρ, ω}, ∀u ∈ ∂Ω_{R}. By (3.13) and (3.25), we have
And so
From (3.6) and (3.26), we get
from which we see that
After multiplying the two sides of the first equation in (3.27) by sin πt and integrating on [0,1], we have
Consequently, we obtain that
It's easy to prove that ω_{0}(t), the solution of LBVF (3.27) satisfies
and accordingly,
by (3.28), we get
Consequently,
We choose
Now, by the additivity of fixed point index, combine (3.23) and (3.31) to conclude that
Therefore, Q has a fixed poind in
From Theorem 3.1, we immediately obtain the following.
Corollary 3.1. Assume (H_{1})(H_{3}) hold, then in each of the following cases:
(i)
the BVP (1.1) has at least one positive solution.
4 Multiple solutions
Next, we study the multiplicity of positive solutions of BVP (1.1) and assume in this section that
(H_{4}) there is a p >0 such that 0 ≤ u ≤ p and 0 ≤ t ≤ 1 imply f(t, u) < ηp, where
(H_{5}) there is a p >0 such that σp ≤ u ≤ p and 0 ≤ t ≤ 1 imply f (t, u) ≥ λp, where
Theorem 4.1. Assume (H_{1})(H_{4}) hold. If
Proof. According to the proof of Theorem 3.1, there exists 0 < r_{0 }< p < R_{1 }<+∞, such that 0 < r < r_{0 }implies i(Q, Ω_{r}, P) = 0 and R ≥ R_{1 }implies i(Q, Ω_{R}, P) = 0.
Next we prove i(Q, Ω_{p}, P) = 1 if (H_{4}) is satisfied. In fact, for every u ∈ ∂Ω_{p}, based on the preceding definition of Q we come to
Consequently,
Therefore, by (ii) of Lemma 2.7 we have
Combined with (3.17), (3.31), and (4.1), we have
Therefore, Q has fixed points u_{1 }and u_{2 }in
means that u_{1}(t) and u_{2}(t) are positive solutions of BVP (1.1) and 0 ≤ u_{1}_{0 }≤ p ≤ u_{2}_{0}. The proof is completed. □
Theorem 4.2. Assume (H_{1})(H_{3}) and (H_{5}) can be established, and
Proof. According to the proof of Theorem 3.1, there exists 0 < ω < p < R_{2 }<+ ∞, such that 0 < r < ω implies i(Q, Ω_{r}, P) = 1 and R ≥ R_{2 }implies i(Q, Ω_{R}, P) = 1.
We now prove that i(Q, Ω_{p}, P) = 0 if (H_{5}) is satisfied. In fact, for every u ∈ ∂Ω_{p}, by (3.13) we come to ρp ≤ ρu_{0 }≤ u(t) ≤ u_{0 }= p, t ∈ [1/4, 3/4], accordingly, by (H_{5}), we have
from the proof of (ii) of Theorem 3.1, we have
Therefore,
Combined with (3.20), (3.23), and (4.2), there exist
Therefore, Q has fixed points u_{1 }and u_{2 }in
Theorem 4.3. Assume that (H_{1})(H_{3}) hold. If
(i) f(t, u) < ηp_{1 }if 0 ≤ t ≤ 1 and 0 ≤ u ≤ p_{1},
(ii) f(t, u) ≥ λp_{2 }if 0 ≤ t ≤ 1 and σp_{2 }≤ u ≤ p_{2},
where η, σ, λ are just as the above, then BVP (1.1) has at least three positive solutions u_{1}, u_{2}, and u_{3 }such that 0 ≤ u_{1}_{0 }≤ p_{1 }≤ u_{2}_{0 }≤ p_{2 }≤ u_{3}_{0}.
Proof. According to the proof of Theorem 3.1, there exists 0 < r_{0 }< p_{1 }< p_{2 }< R_{3 }<+∞, such that 0 < r < r_{0 }implies i(Q, Ω_{r}, P) = 0 and R ≥ R_{3 }implies i(Q, Ω_{R}, P) = 1.
From the proof of Theorems 4.1 and 4.2, we have
Therefore, Q has fixed points u_{1}, u_{2 }and u_{3 }in
Competing interests
The author declares that they have no competing interests.
Authors' contributions
WL conceived of the study, and participated in its design and coordination. The author read and approved the final manuscript.
Acknowledgements
The author is very grateful to the anonymous referees for their valuable suggestions, and to be sponsored by the Tutorial Scientific Research Program Foundation of Education Department of Gansu Province P.R.China(111005).
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